已知0<X1<3,Xn=根号下Xn-1(3-Xn-1)证明{Xn}极限存在,并求极限
问题描述:
已知0<X1<3,Xn=根号下Xn-1(3-Xn-1)证明{Xn}极限存在,并求极限
答
证明:因为0<x1<3所以x(n+1)<=[xn+(3-xn)]/2=3/2所以{xn}有界又x(n+1)=√[Xn(3-Xn)]>=√[Xn(3-3/2)]=√(3/2)xn>=xn所以{xn}递增单调有界数列必有极限,设x=limxn=limx(n+1),则x=√x(3-x)解得x=3...